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Existence and characterization of optimal dynamic pricing strategies with reference-price effects

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Abstract

In this paper, we revisit the problem of existence of an optimal dynamic pricing strategy when the demand depends on a reference price. Using alternative assumptions and technique to those typically used in the literature, we show the existence of a unique pricing policy parametrized in the initial value of the reference price. Our main results are as follows: if this value is low enough, then the best option is to implement a penetration pricing. If it is high enough, then price skimming is optimal. If the initial reference price has an intermediate value, then constant pricing throughout the planning horizon is optimal.

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Notes

  1. Sorger (1988) addressed the local stability of optimal behavior when the firm’s choices include pricing and advertising policies.

  2. Assuming that the price chosen is the largest maximizers ensures that the pricing policy is a well-defined function.

  3. Our framework is less general than the one of Popescu and Wu (2007).

  4. The proof that when the initial value of the reference price takes intermediate values it is optimal to set the price equal to the reference price relies on a clever argument proposed by Fibich et al. (2003).

  5. The impact of the presence of reference price on the profitability of price promotion is analyzed in Greenleaf (1995).

  6. See Wang (2016) on intertemporal price discrimination through reference price.

  7. Of course, it is never optimal to always set \(p_{t}\) equal to its maximal value.

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Correspondence to Georges Zaccour.

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Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

We would like to thank the two Reviewers for their helpful comments. The third author’s research is supported by NSERC Canada, Grant RGPIN-2016-04975. He also thanks CRED, Université Panthéon-Assas Paris II, for its hospitality.

Appendix: Proofs

Appendix: Proofs

Proof of Lemma 1

Using the dynamics in (4) and the established bound on price \(p_{t}\) in (6), we have

$$\begin{aligned} q_{t+1}= & {} q_{t}+s(p_{t}-q_{t})=(1-s)q_{t}+sp_{t}, \\\le & {} (1-s)q_{t}+\frac{s\lambda }{\theta +\phi _{1}}+\frac{s\phi _{2}}{ \theta +\phi _{2}}q_{t}, \\= & {} q_{t}\left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}}\right) +\frac{ s\lambda }{\theta +\phi _{1}} \\\le & {} \left( q_{t-1}\left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}}\right) +\frac{ s\lambda }{\theta +\phi _{1}}\right) \left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}} \right) +\frac{s\lambda }{\theta +\phi _{1}} \\= & {} q_{t-1}\left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}}\right) ^{2}+\frac{ s\lambda }{\theta +\phi _{1}}\left( 1+\left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}}\right) \right) \\&\ldots \\\le & {} q_{0}\left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}}\right) ^{t+1}+ \frac{s\lambda }{\theta +\phi _{1}}\sum _{k=0}^{t}\left( 1-s+\frac{s\phi _{2} }{\theta +\phi _{2}}\right) ^{k} \\= & {} q_{0}\left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}}\right) ^{t+1}+\frac{ s\lambda }{\theta +\phi _{1}}\left( \frac{1-\left( 1-s+\frac{s\phi _{2}}{ \theta +\phi _{2}}\right) ^{t}}{1-\left( 1-s+\frac{s\phi _{2}}{\theta +\phi _{2}}\right) }\right) , \\< & {} q_{0}+\frac{s\lambda }{\theta +\phi _{1}}\frac{1}{1-\left( 1-s+\frac{ s\phi _{2}}{\theta +\phi _{2}}\right) }, \\= & {} q_{0}+\frac{\lambda (\theta +\phi _{2})}{\theta +\phi _{1}+\phi _{2}} \equiv \bar{{\bar{q}}}(q_{0}). \end{aligned}$$

\(\square \)

Proof of Lemma 2

Let a sequence \(({\bar{p}}^{n},{\bar{q}}^{n})\) in \({\mathcal {A}}(k_{0})\) converges to \(({\bar{p}},{\bar{q}})\). By construction, \(p_{t}^{n}D(p_{t}^{n},q_{t}^{n}) \le {\bar{\Pi }}(\bar{{\bar{q}}}(q_{0}))\). Since \((p_{t},q_{t})\) is upper-bounded and \(\beta <1,\) we can assert that for all \(\epsilon >0\), there exists \(T(\epsilon )\) such that for all \(T\ge T(\epsilon )\),

$$\begin{aligned} \sum _{t=T(\epsilon )}^{+\infty }\beta ^{t}{\bar{\Pi }}(\bar{{\bar{q}}} (q_{0}).)<\epsilon . \end{aligned}$$

But this implies that for all n

$$\begin{aligned} \sum _{t=0}^{+\infty }\beta ^{t}p_{t}^{n}D(p_{t}^{n},q_{t}^{n})<\epsilon . \end{aligned}$$

Therefore, we have for all n, and for all \(T\ge T(\epsilon )\)

$$\begin{aligned} {\mathcal {F}}({\bar{p}}^{n},{\bar{q}}^{n})=\sum _{t=0}^{+\infty }\beta ^{t}p_{t}^{n}D(p_{t}^{n},q_{t}^{n})\le \sum _{t=0}^{T}\beta ^{t}p_{t}^{n}D(p_{t}^{n},q_{t}^{n})+\epsilon . \end{aligned}$$

Consequently, we have

$$\begin{aligned} \limsup _{n\rightarrow +\infty }{\mathcal {F}}({\bar{p}}^{n},{\bar{q}}^{n})\le \sum _{t=0}^{T}\beta ^{t}p_{t}^{n}D(p_{t},q_{t})+\epsilon , \end{aligned}$$

because \(D(p_{t},q_{t})\) is continuous. Since the above inequality is true for all \(T\ge T(\epsilon )\), and for all \(\epsilon >0\), it follows that:

$$\begin{aligned} \limsup _{n\rightarrow +\infty }{\mathcal {F}}({\bar{p}}^{n},{\bar{q}}^{n})\le {\mathcal {F}}({\bar{p}},{\bar{q}}). \end{aligned}$$

\(\square \)

Proof of Proposition 2

Let \(q_{0}\) and \(q_{0}^{\prime }\) be two non negative real numbers such that \(q_{0}<q_{0}^{\prime }\). Let \((p_{t})_{t}\) be any feasible sequence of prices (namely, \(0\le p_{t}\le \bar{{\bar{p}}}(q_{0})\) for all t). Now, let \((q_{t})_{t}\) be the sequence of reference prices defined by

$$\begin{aligned} q_{t+1}=(1-s)q_{t}+sp_{t}, \end{aligned}$$

whose first-term is \(q_{0}\). Define by \((q_{t}^{\prime })_{t}\) the sequence of reference prices obtained as follows:

$$\begin{aligned} q_{1}^{\prime }&=(1-s)q_{0}^{\prime }+sp_{0}, \end{aligned}$$
(10)
$$\begin{aligned} q_{2}^{\prime }&=(1-s)q_{1}^{\prime }+sp_{1}, \end{aligned}$$
(11)
$$\begin{aligned} \cdots&=\cdots \end{aligned}$$
(12)
$$\begin{aligned} q_{t+1}^{\prime }&=(1-s)q_{t}^{\prime }+sp_{t}. \end{aligned}$$
(13)

Clearly, we have \(q_{t}^{\prime }>q_{t}\) for all t. Assume now that the sequence \((p_{t},q_{t})_{t}\) maximizes \({\mathcal {F}}({\underline{p}},\underline{ q})\), i.e., \(V(q_{0})={\mathcal {F}}({\underline{p}},{\underline{q}})\). Then, we have

$$\begin{aligned} V(q_{0})&=\sum _{t=0}^{+\infty }\beta ^{t}D(p_{t},q_{t}) \end{aligned}$$
(14)
$$\begin{aligned}&<\sum _{t=0}^{+\infty }\beta ^{t}D(p_{t},q_{t}^{\prime }) \end{aligned}$$
(15)
$$\begin{aligned}&\le V(q_{0}^{\prime }), \end{aligned}$$
(16)

where the last line follows from the definition of the value function. \(\square \)

Proof of Proposition 5

First, use Eq. (4) to express \(p_{t}\) as a function of \(q_{t}\) and \(q_{t+1}\), namely

$$\begin{aligned} p_{t}=\frac{q_{t+1}-(1-s)q_{t}}{s}. \end{aligned}$$

As \(p_{t}\) must be non-negative, we thus have \(q_{t+1}\ge (1-s)q_{t}\). Moreover, \(p_{t}\) is upper bounded by the condition \(D(p_{t},q_{t})\ge 0\). Therefore, we must have

$$\begin{aligned} q_{t+1}\le \frac{s\lambda +(\theta (1-s)+\phi _{1})q_{t}}{\theta +\phi _{1}} . \end{aligned}$$

Observe that it is never optimal to set \(p_{t}=0\) at a date t. That is because, setting \(p_{t}=\epsilon \) arbitrarily small enables to obtain positive revenues, and to increase \(q_{t+1}\). But since the value function is increasing (by the same argument as that used previously) in q, it is immediate that setting \(p_{t}=0\) is never an optimal choice (and therefore \( q_{t+1}>(1-s)q_{t}\)).

Now, assume that there exists a date t such that \(p_{t}\) is equal to its maximal value, that is, \(p_{t}=\frac{\lambda +\phi _{1}q_{t}}{\theta +\phi _{1}}\), and \(p_{t+1}\) is lower than its maximal value.Footnote 7 We shall show that this cannot be an optimal choice.

Suppose that \((p_{t},q_{t})_{t}\) is an optimal process such that there exists a date t where \(p_{t}\) is equal to its maximal value and \(p_{t+1}\) is lower than its maximal value. Let \(\epsilon >0\) be such that \( p_{t}-\epsilon >0\). Construct a process \(({\tilde{p}}_{t},{\tilde{q}}_{t})_{t}\) in the following way.

$$\begin{aligned}&\left. \begin{array}{c} {\tilde{p}}_{j}=p_{j}\\ {\tilde{q}}_{j}=q_{j} \end{array} \right\} ,\;\;for\;j=1,\ldots ,t-1, \\&{\tilde{p}}_{t} =p_{t}-\epsilon , \\&{\tilde{q}}_{t} =q_{t}+s\epsilon , \\&{\tilde{p}}_{t+1} =p_{t+1}, \\&{\tilde{q}}_{t+1} =s{\tilde{p}}_{t}+(1-s){\tilde{q}}_{t}, \\&\left. \begin{array}{c} {\tilde{p}}_{j}=p_{j}\\ {\tilde{q}}_{j}=q_{j} \end{array} \right\} ,\;\;for\;j>t+1. \end{aligned}$$

Notice that \(({\tilde{p}}_{t},{\tilde{q}}_{t})_{t}\) is an admissible process.

Now,

$$\begin{aligned}&\sum _{t=0}^{+\infty }\beta ^{t}{\tilde{p}}_{t}D({\tilde{p}}_{t},{\tilde{q}} _{t})-\sum _{t=0}^{+\infty }\beta ^{t}p_{t}D(p_{t},q_{t}) \\&\quad =\beta ^{t}{\tilde{p}}_{t}D({\tilde{p}}_{t},{\tilde{q}}_{t})-\beta ^{t}p_{t}D(p_{t},q_{t})+\beta ^{t+1}{\tilde{p}}_{t+1}D({\tilde{p}}_{t+1},\tilde{q }_{t+1})\\&\qquad -\,\beta ^{t+1}p_{t+1}D(p_{t+1},q_{t+1})\\&\quad =\beta ^{t}{\tilde{p}}_{t}D({\tilde{p}}_{t},{\tilde{q}}_{t})-\beta ^{t}p_{t}D(p_{t},{\tilde{q}}_{t})+\beta ^{t}p_{t}\left( D(p_{t},{\tilde{q}} _{t})-D(p_{t},q_{t})\right) \\&\qquad +\,\beta ^{t+1}{\tilde{p}}_{t+1}\left( D({\tilde{p}}_{t+1},{\tilde{q}}_{t+1})-D( {\tilde{p}}_{t+1},q_{t+1})\right) \\&\quad \ge \beta ^{t}(-\epsilon )(\lambda -2(\theta +\phi _{1}){\tilde{p}} _{t}+\phi _{1}{\tilde{q}}_{t})+\beta ^{t}p_{t}\phi _{1}(s\epsilon )+\beta ^{t+1}{\tilde{p}}_{t+1}\phi _{1}(-s^{2}\epsilon ). \end{aligned}$$

The last inequality follows from the concavity of the profit function \( p_{t}D(p_{t},q_{t})\) with respect to \(p_{t}\). Hence,

$$\begin{aligned}&\sum _{t=0}^{+\infty }\beta ^{t}{\tilde{p}}_{t}D({\tilde{p}}_{t},{\tilde{q}} _{t})-\sum _{t=0}^{+\infty }\beta ^{t}p_{t}D(p_{t},q_{t}) \\&\quad \ge \beta ^{t}[(-\epsilon )(\lambda -2(\theta +\phi _{1}){\tilde{p}} _{t}+\phi _{1}{\tilde{q}}_{t})+p_{t}\phi _{1}(s\epsilon )+\beta p_{t+1}\phi _{1}(-s^{2}\epsilon )]\\&\quad =\beta ^{t}[(-\epsilon )(\lambda -2(\theta +\phi _{1})(p_{t}-\epsilon )+\phi _{1}(q_{t}+s\epsilon ))+p_{t}\phi _{1}(s\epsilon )+\beta p_{t+1}\phi _{1}(-s^{2}\epsilon )]. \end{aligned}$$

The sign of the above last expression is the same as the sign of

$$\begin{aligned}&(-\lambda +2(\theta +\phi _{1})(p_{t}-\epsilon )-\phi _{1}(q_{t}+s\epsilon ))+p_{t}\phi _{1}s-\beta p_{t+1}\phi _{1}s^{2}\\&\quad =(-\lambda +2(\theta +\phi _{1})\left( \frac{\lambda +\phi _{1}q_{t}}{ \theta +\phi _{1}}-\epsilon \right) -\phi _{1}(q_{t}+s\epsilon ))+p_{t}\phi _{1}s-\beta p_{t+1}\phi _{1}s^{2} \end{aligned}$$

When \(\epsilon \) goes to zero, we have

$$\begin{aligned}&(-\lambda +2(\theta +\phi _{1})\left( \frac{\lambda +\phi _{1}q_{t}}{ \theta +\phi _{1}}-\epsilon \right) -\phi _{1}(q_{t}+s\epsilon ))+p_{t}\phi _{1}s-\beta p_{t+1}\phi _{1}s^{2} \\&\quad =(\lambda +\phi _{1}q_{t})+p_{t}\phi _{1}s-\beta p_{t+1}\phi _{1}s^{2}>0. \end{aligned}$$

Indeed, since

$$\begin{aligned} p_{t+1}<\frac{\lambda +\phi _{1}q_{t+1}}{\theta +\phi _{1}}=\frac{\lambda +\phi _{1}(sp_{t}+(1-s)q_{t})}{\theta +\phi _{1}}, \end{aligned}$$

we have

$$\begin{aligned} \beta p_{t+1}\phi _{1}s^{2}<({\theta +\phi _{1}})p_{t+1}<\lambda +\phi _{1}(sp_{t}+(1-s)q_{t})<\lambda +\phi _{1}(sp_{t}+q_{t}). \end{aligned}$$

Therefore, \((p_{t},q_{t})_{t}\) cannot be optimal. \(\square \)

Proof of Proposition 7

Clearly, from Lemma 3, we know that the value of the problem (8) is no lower than the value of our original problem. From Proposition (5), we know that this problem has an interior solution.

Now, any interior solution must satisfy the following Euler condition:

$$\begin{aligned}&\frac{1}{s}\left( \lambda -\theta \left( \frac{q_{t+1}-(1-s)q_{t}}{s}\right) -\phi _{1}\left( \frac{q_{t+1}-q_{t}}{s}\right) \right) \\&\quad -\,\frac{1}{s}\left( \frac{q_{t+1}-(1-s)q_{t}}{s}\right) (\theta +\phi _{1}) \\&\quad +\,\beta \left\{ -\frac{(1-s)}{s}\left( \lambda -\theta \left( \frac{ q_{t+2}-(1-s)q_{t+1}}{s}\right) -\phi _{1}\left( \frac{q_{t+2}-q_{t+1}}{s} \right) \right) \right. \\&\quad \left. +\left( \frac{q_{t+2}-(1-s)q_{t+1}}{s}\right) \frac{\left( (1-s)\theta +\phi _{1}\right) }{s}\right\} =0, \end{aligned}$$

which can be reduced, after straightforward manipulations, to

$$\begin{aligned}&q_{t+2}-2q_{t+1}\frac{\left( \theta +\phi _{1}+(1-s)\beta (\theta (1-s)+\phi _{1})\right) }{\beta \left( 2(1-s)(\theta +\phi _{1})+s\phi _{1}\right) } \nonumber \\&\quad +\,\frac{q_{t}}{\beta }+s\lambda \frac{1-\beta (1-s)}{\beta \left( 2(1-s)(\theta +\phi _{1})+s\phi _{1}\right) }=0. \end{aligned}$$
(17)

The above equation allows to define the following characteristic polynomial:

$$\begin{aligned} P(\mu )=\mu ^{2}-2\mu \frac{\left( \theta +\phi _{1}+(1-s)\beta (\theta (1-s)+\phi _{1})\right) }{\beta \left( 2(1-s)(\theta +\phi _{1})+s\phi _{1}\right) }+\frac{1}{\beta }=0. \end{aligned}$$

Notice that after little algebra we can show that \(P(1)<0\). Therefore, the polynomial has two positive roots, one lower than one (call it \(\mu _{1}\)) and the other strictly higher than one (\(\mu _{2}\)). The general solution reads

$$\begin{aligned} q_{t}=\rho _{1}\mu _{1}^{t}+\rho _{2}\mu _{2}^{t}+{\overline{q}}, \end{aligned}$$
(18)

where

$$\begin{aligned} {\overline{q}}&=\frac{s\lambda \big (1-\beta (1-s)\big )}{2(\theta +\phi _{1})+2(1-s)\beta \big (\theta (1-s)+\phi _{1}\big )-(1+\beta )\big ( 2(1-s)(\theta +\phi _{1})+s\phi _{1}\big )}, \\&=\frac{\left( 1-\beta (1-s)\right) \lambda }{2\theta (1-\beta )+2s\beta \theta +(1-\beta )\phi _{1}} \end{aligned}$$

Next, we need to determine the values of \(\rho _{1}\) and \(\rho _{2}\) in (18).

We must have \(\rho _{2}=0\); otherwise, the value of \(q_{t}\) would grow without limit, which is not feasible (this can be proved by an argument similar to the one used before, see Section 3). It is then easy to see that \( \rho _{1}=q_{0}-{\bar{q}}\), and thus that

$$\begin{aligned} q_{t}={\overline{q}}(1-\mu _{1}^{t})+q_{0}\mu _{1}^{t}. \end{aligned}$$
(19)

This expression gives the solution to our problem, since it generates the unique bounded solution to the Euler equation.

Finally, since \(p_{t}=\frac{q_{t+1}-(1-s)q_{t}}{s}\), we see that

$$\begin{aligned} p_{t}\le q_{t},\,\,\forall t,\;\;\text {if, and only if},\;\;q_{t+1}\le q_{t}, \end{aligned}$$

which is equivalent to \({\overline{q}}\le q_{0}\), see (19), which holds by assumption. The result follows from Proposition 3. \(\square \)

Proof of Proposition 9

The proof follows the astute argument in Fibich et al. (2003) in the continuous-time case. Consider first the problem with \(\phi _{1}=\phi _{2}=\phi \). Using the same argument as in the proof of Proposition 7, we can prove that the optimal solution is given by

$$\begin{aligned} q_{t}={q}(\phi )(1-\mu ^{t})+q_{0}\mu ^{t}={q}(\phi )+\mu ^{t}(q_{0}-{q} (\phi )), \end{aligned}$$

where \(0<\mu <1\) and \(q(\phi )=\frac{\left( 1-\beta (1-s)\right) \lambda }{ 2\theta (1-\beta )+2s\beta \theta +(1-\beta )\phi }\).

It is immediate to see that when \(q_{0}={q}(\phi )\), then \(q_{t}=q(\phi )\) for all t is an optimal solution for the problem with \(\phi _{1}=\phi _{2}=\phi \).

Now, It is clear that the function \(q(\phi )=\frac{\left( 1-\beta (1-s)\right) \lambda }{2\theta (1-\beta )+2s\beta \theta +(1-\beta )\phi }\) is decreasing in \(\phi \). Let \(\phi ^{*}\) satisfy \(q_{0}=\frac{\left( 1-\beta (1-s)\right) \lambda }{2\theta (1-\beta )+2s\beta \theta +(1-\beta )\phi ^{*}}\). Therefore, \(\phi _{1}\le \phi ^{*}\le \phi _{2}\) since \({\underline{q}}=q(\phi _{2})\le q_{0}\le {\bar{q}}=q(\phi _{1})\).

Set

$$\begin{aligned}&g(\phi _{1},\phi _{2},(q_{t})_{t})=\sum _{t=0}^{+\infty }\beta ^{t}p_{t}D(p_{t},q_{t})\\&\quad =\sum _{t=0}^{+\infty }\beta ^{t}\left( \frac{ q_{t+1}-(1-s)q_{t}}{s}\right) D\left( \left( \frac{q_{t+1}-(1-s)q_{t}}{s} \right) ,q_{t}\right) \end{aligned}$$

and \(G(\phi _{1},\phi _{2})=\max _{q_{t}}g(\phi _{1},\phi _{2},(q_{t})_{t})\) under (1) and (2).

It is easy to see that g is increasing in \(\phi _{1}\) and decreasing in to \(\phi _{2}\). Consequently, we have \(g(\phi _{1},\phi _{2},(q_{t})_{t})\le g(\phi ^{*},\phi ^{*},(q_{t})_{t})\), and

$$\begin{aligned} g(\phi _{1},\phi _{2},(q_{0})_{t})\le G(\phi _{1},\phi _{2})\le G(\phi ^{*},\phi ^{*}). \end{aligned}$$

But

$$\begin{aligned} g(\phi _{1},\phi _{2},(q_{0})_{t})=g(\phi ^{*},\phi ^{*},(q_{0})_{t})=G(\phi ^{*},\phi ^{*}), \end{aligned}$$

since when \(q_{t}=q_{0}\) we have \(p_{t}=q_{0}\) and the problem does no more depend on \(\phi _{1}\) and \(\phi _{2}\).

Finally, we get

$$\begin{aligned} G(\phi ^{*},\phi ^{*})\le g(\phi _{1},\phi _{2},(q_{0})_{t})\le G(\phi _{1},\phi _{2})\le G(\phi ^{*},\phi ^{*}). \end{aligned}$$

Therefore, \(g(\phi _{1},\phi _{2},(q_{0})_{t})=G(\phi _{1},\phi _{2}),\)which means that \(q_{t}=q_{0}\) for all t is an optimal solution. \(\square \)

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Crettez, B., Hayek, N. & Zaccour, G. Existence and characterization of optimal dynamic pricing strategies with reference-price effects. Cent Eur J Oper Res 28, 441–459 (2020). https://doi.org/10.1007/s10100-019-00645-w

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